Why does a d-polytope being k-neighborly for 1 <= k <= d imply the polytope is neighborly for all k' such that 1 <= k' <= k?

Question

I'm reading Grunbaum's Convex Polytopes where he cites the following theorem in a proof by contradiction for a larger theorem:

"If $P$ is a $k$-neighborly $d$-polytope, and if $1 \le k' \le k$, then $P$ is $k'$-neighborly."

The preceding text notes that this is a result of the vertices of $P$ being affinely independent and the consequence that

"The convex hull of every $k$ members of $vert(P)$ is a $(k-1)$-face of $P$; more precisely, it is a $(k-1)$-simplex. On the other hand, since each $(k-1)$-face of $P$ contains at least $k$ vertices, it follows that every $(k-1)$-face of $P$ is a $(k-1)$-simplex."

How does this connection work? Does it have to do with the faces of $P$ being simplices, so polytopes themselves, and $k$-neighborliness somehow transfers?

I'm sorry if this isn't clear; I think I have some intuitive idea of what's going on but I'm not sure how to express it.

Answer 1

Just cf. to https://en.wikipedia.org/wiki/Neighborly_polytope . There the second paragraphe states:

In a k-neighborly polytope with k ≥ 3, every 2-face must be a triangle, and in a k-neighborly polytope with k ≥ 4, every 3-face must be a tetrahedron. More generally, in any k-neighborly polytope, all faces of dimension less than k are simplices.

And all faces of simplices surely are lower dimensional simplices in turn. Then just put these 2 statements together.

--- rk